Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct TreeNode {
  int val;
  TreeNode* left;
  TreeNode* right;
  TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
struct ListNode {
  int val;
  ListNode* next;
  ListNode(int x) : val(x), next(NULL) {}
};
int nxt() {
  int x;
  cin >> x;
  return x;
}
long long nxtLong() {
  long long x;
  cin >> x;
  return x;
}
string nxtStr() {
  string x;
  cin >> x;
  return x;
}
char nxtChar() {
  char x;
  cin >> x;
  return x;
}
double nxtDouble() {
  double x;
  cin >> x;
  return x;
}
long double nxtLDouble() {
  long double x;
  cin >> x;
  return x;
}
TreeNode* makeTree(int n) {
  queue<TreeNode*> tree_nodes;
  TreeNode* root;
  if (n) {
    root = new TreeNode(nxt());
    tree_nodes.push(root);
  } else {
    return NULL;
  }
  while (!tree_nodes.empty()) {
    TreeNode* node = tree_nodes.front();
    tree_nodes.pop();
    bool left;
    cin >> left;
    if (left) {
      node->left = new TreeNode(nxt());
      tree_nodes.push(node->left);
    }
    bool right;
    cin >> right;
    if (right) {
      node->right = new TreeNode(nxt());
      tree_nodes.push(node->right);
    }
  }
  return root;
}
ListNode* makeLinkedList(int n) {
  ListNode* list = new ListNode(0);
  ListNode* dummy = list;
  for (int i = 0; i < n; ++i) {
    list->next = new ListNode(nxt());
    list = list->next;
  }
  return dummy->next;
}
void solve() {
  int n = nxt(), m = nxt();
  vector<vector<int>> in(n, vector<int>(m));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cin >> in[i][j];
    }
  }
  vector<int> row(n, INT_MAX);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      row[i] = min(row[i], in[i][j]);
    }
  }
  vector<int> col(m, INT_MAX);
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j < n; ++j) {
      col[i] = min(col[i], in[j][i]);
    }
  }
  vector<vector<int>> ans(n, vector<int>(m));
  int cnt = 0;
  vector<pair<string, int>> order;
  for (int i = 0; i < n; ++i) {
    int mi = row[i];
    int mini = INT_MAX;
    for (int j = 0; j < m; ++j) {
      mini = min(mini, ans[i][j]);
    }
    if (mi && mi > mini) {
      cnt += mi - mini;
      for (int j = 0; j < mi - mini; ++j) {
        order.push_back({"row", i + 1});
      }
      for (int j = 0; j < m; ++j) {
        ans[i][j] += mi - mini;
      }
    }
  }
  for (int i = 0; i < m; ++i) {
    int mi = col[i];
    int mini = INT_MAX;
    for (int j = 0; j < n; ++j) {
      mini = min(mini, ans[j][i]);
    }
    if (mi && mi > mini) {
      cnt += mi - mini;
      for (int j = 0; j < mi - mini; ++j) {
        order.push_back({"col", i + 1});
      }
      for (int j = 0; j < n; ++j) {
        ans[j][i] += mi - mini;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (ans[i][j] != in[i][j]) {
        cout << -1 << endl;
        return;
      }
    }
  }
  int cnt2 = 0;
  vector<pair<string, int>> order2;
  vector<vector<int>> ans2(n, vector<int>(m));
  for (int i = 0; i < m; ++i) {
    int mi = col[i];
    int mini = INT_MAX;
    for (int j = 0; j < n; ++j) {
      mini = min(mini, ans2[j][i]);
    }
    if (mi && mi > mini) {
      cnt2 += mi - mini;
      for (int j = 0; j < mi - mini; ++j) {
        order2.push_back({"col", i + 1});
      }
      for (int j = 0; j < n; ++j) {
        ans2[j][i] += mi - mini;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    int mi = row[i];
    int mini = INT_MAX;
    for (int j = 0; j < m; ++j) {
      mini = min(mini, ans2[i][j]);
    }
    if (mi && mi > mini) {
      cnt2 += mi - mini;
      for (int j = 0; j < mi - mini; ++j) {
        order2.push_back({"row", i + 1});
      }
      for (int j = 0; j < m; ++j) {
        ans2[i][j] += mi - mini;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (ans2[i][j] != in[i][j]) {
        cout << -1 << endl;
        return;
      }
    }
  }
  if (cnt <= cnt2) {
    cout << cnt << endl;
    for (auto pr : order) {
      cout << pr.first << ' ' << pr.second << endl;
    }
  } else {
    cout << cnt2 << endl;
    for (auto pr : order2) {
      cout << pr.first << ' ' << pr.second << endl;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(9);
  solve();
  return 0;
}